Question

logₑ(100.1),Find approximate value.

Answer 1

we have to find approximate value of $log_e(100.1)$

we have to use just logrithmn concepts .
like, $log_e(10)=2.302$, $log_eMN=log_eM+log_eN$,
$log_eA^n=nlog_eA$


now, $log_e(100.1)$

= $log_e(100+0.1)$

= $log_e\{100(1+0.001)\}$

= $log_e100+log_e(1+0.001)$

= $log_e10^2+log_e(1+0.001)$

= $2log_e(10)+log_e(1+0.001)$

now , we have to use expansion of log(1+0.001)
from Taylor's series ,
log(1 + x) = x - x²/2 + x³/3 - x⁴/4 .......
here, x = 0.001 = 10^-3

now, log(1 + 0.001)=10^-3 - (10^-3)²/2 + (10^-3)³/3 ...
≈ 10^-3 - 0.5(10)^-6
≈ 0.001 - 0.5 × 0.000001
≈ 0.001 - 0.0000005
≈ 0.0009995

= $2\times2.302+0.0009995$

= 4.604 + 0.0009995
= 4.6049995

hence, approximate value of logₑ(100.1)≈ 4.6049995

Answer 2

Dear Student:

For finding we will define x =100

and Δx=0.1

and use,Δy=dy/dx*(Δx)

And,also Δy=f(x+Δx)-f(x)

See the attachment: